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coordinate geometry

by vscid » Sat Aug 01, 2009 1:33 pm
On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

1√2
1
√2
√3
2√3


Please explain.
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by tom4lax » Tue Aug 04, 2009 5:02 am
Am I missing something, or do those coordinates result in a non-square quad.?
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Re: coordinate geometry

by shahdevine » Tue Aug 04, 2009 1:56 pm
vscid wrote:On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

1√2
1
√2
√3
2√3


Please explain.
----------------
please double check coordinates. they create non-square.
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Re: coordinate geometry

by ankitns » Tue Aug 04, 2009 2:37 pm
It will be a square where the sides are NOT parallel to either of the axes.

Cheers.
shahdevine wrote:
vscid wrote:On the coordinate plane (6, 2) and (0, 6) are the endpoints of the diagonal of a square. What is the distance between point (0, 0) and the closest vertex of the square?

1√2
1
√2
√3
2√3


Please explain.
----------------
please double check coordinates. they create non-square.
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by italian7745 » Wed Aug 05, 2009 3:01 am
OA is C...

Cant we make a square with sides (0,6) , (1,1) , (6,2) and (5,7)...???
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by tom4lax » Wed Aug 05, 2009 4:39 am
Would you mind explaining how you got to those coordinates?
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by vscid » Wed Aug 05, 2009 5:22 pm
tom4lax wrote:Am I missing something, or do those coordinates result in a non-square quad.?
Why do you think the coordinates result in a non-square?
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by tom4lax » Wed Aug 05, 2009 6:53 pm
endpoints of diagonal (0,6) and (6,2) = difference in x values of 6 and y values of 4, sides aren't of same length. However, if the sides aren't parallel to the axis, then possibly it is a square. I just don't know how you would find these points?
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by aspiregmat » Thu Aug 06, 2009 2:11 am
Don't know a quick way to find.. It took me almost 10 min to reach to the solution...


The two points are (6,2) and (0,6)

The length of diagonal = sqrt(52)

Length of the side = sqrt(26)

Now the two points we know and the length we know so using the formula sqrt( sqr(x1-x2) + sqr(y1-y2)) we can get two equations..

Let the 3rd point in square be (x,y)

For (6,2) == x^2 + y^2 -12x - 4y + 14 = 0 ------- 1

for (6,0) == x^2 + y^2 -12y + 10 = 0 ------------- 2

subtract one from other then we get 3x - 2y = 1

x = 1/3(1+2y)

Sub x in 2

we get y^2 - 8y +7 = 0

so values of Y = 1 , 7
then value of X = 1, 5

(1,1) is nearer to (0,0)

so the distance = sqrt(2)
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by vscid » Thu Aug 06, 2009 4:41 am
tom4lax wrote:endpoints of diagonal (0,6) and (6,2) = difference in x values of 6 and y values of 4, sides aren't of same length. However, if the sides aren't parallel to the axis, then possibly it is a square. I just don't know how you would find these points?
Nowhere it is mentioned that the sides are parallel to the axis.
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by tom4lax » Thu Aug 06, 2009 4:42 am
Right, thanks, I realize that now.

How did you solve the question?
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by vscid » Thu Aug 06, 2009 5:48 am
tom4lax wrote:Right, thanks, I realize that now.

How did you solve the question?
Almost the same way, aspireGMAT did.
The GMAT is indeed adaptable. Whenever I answer RC, it proficiently 'adapts' itself to mark my 'right' answer 'wrong'.
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